Question

Find the zeroes of the quadratic polynomial y square + 92 y + 1920​

Answer 1

answe

y²+92y+1920=0

y²+60y+32y+1920=0

y(y+60)+32(y+60)=0

y(60) (y+32)

Answer 2

Solution :

Given :

y² + 92y + 1920

$\bf{\underline{\bf{Explanation\::}}}}$

By using quadratic formula :

As the given polynomial compared with ax² + bx + c

• a = 1
• b = 92
• c = 1920

Now;

$\longrightarrow\tt{x=\dfrac{-b\pm\sqrt{b^{2}-4ac } }{2a} }\\\\\\\longrightarrow\tt{x=\dfrac{-92\pm\sqrt{(92)^{2} -4\times 1\times 1920} }{2\times 1} }\\\\\\\longrightarrow\tt{x=\dfrac{-92\pm\sqrt{8464-7680} }{2} }\\\\\\\longrightarrow\tt{x=\dfrac{-92\pm\sqrt{784} }{2} }\\\\\\\longrightarrow\tt{x=\dfrac{-92\pm28}{2} }\\\\\\\longrightarrow\tt{x=\dfrac{-92+28}{2} \:\:Or\:\:x=\dfrac{-92-28}{2} }\\\\\\\longrightarrow\tt{x=\cancel{\dfrac{-64}{2}} \:\:Or\:\:x=\cancel{\dfrac{-120}{2} }}$

$\longrightarrow\bf{x=-32\:\:\:Or\:\:\:x=-60}}$

Thus;

Zeroes of the polynomial will be -32 and -60 .